Question

是否有办法以流利的方式证明一种以上财产无效?

例如,不使用挥发器,可能可以通过实施四甲苯胺。

    public class Request : IValidatableObject
    {
        public int Id { get; init; }

        public string Property1 { get; init; }
        public string Property2 { get; init; }
        public string Property3 { get; init; }

        public IEnumerable<ValidationResult> Validate(ValidationContext validationContext)
        {
            if (Property1 == null && Property2 == null && Property3 == null)
            {
                yield return new ValidationResult("Some message");
            }
        }
    }

我找到了通过压倒一切的Validate使用流放器的方法,但这并不是为了产生挥发性变压器。

    public class RequestValidator : AbstractValidator<Request>
    {
        public override FluentValidation.Results.ValidationResult Validate(ValidationContext<Request> context)
        {
            if (context.InstanceToValidate.Property1 is null &&
                context.InstanceToValidate.Property2 is null &&
                context.InstanceToValidate.Property3 is null)
            {
                return new FluentValidation.Results.ValidationResult(new[]
                {
                    new ValidationFailure("", "Custom message")
                });
            }
        
            return new FluentValidation.Results.ValidationResult();
        }
    }

Answer 1

One possible solution:

public class RequestValidator : AbstractValidator<Request>
{
    public RequestValidator()
    {
        RuleFor(x => x)
            .Must(
                x => x.Property1 is not null
                  || x.Property2 is not null
                  || x.Property3 is not null)
            .WithMessage("Custom message");
    }
}

Answer 2

我发现,这样做的另一个方式/方式是:

// I have nullable enabled

public class Request
{
    public int Id { get; init; }

    public string? Property1 { get; init; }
    public string? Property2 { get; init; }
    public string? Property3 { get; init; }
}

public class RequestValidator : AbstractValidator<Request>
{
    public RequestValidator()
    {
        RuleFor(x => new List<string?> { x.Property1, x.Property2, x.Property3 })
            .Must(x => x.Any(x => x is not null))
            .WithMessage("You need to have at least one of Property 1, 2 or 3.");
    }
}

如果它们重新提出反对,而不是原始类型,那么你可以做这样的事情。

RuleFor(x => new List<object?> { x.Property1, x.Property2, ... })
    ...

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